Work and Rotational Kinetic Energy — Question 4

Question 4

A fixed hinge does no work while a rod falls

A uniform rod of mass $2\,\mathrm{kg}$ and length $1.0\,\mathrm m$ is hinged at one end and released from rest horizontally. The hinge is frictionless. Use $I=ML^2/3$ and $g=9.8\,\mathrm{m/s^2}$.

1. Find angular speed when the rod first reaches the downward vertical.

2. If a hinge brake instead removes half the released gravitational energy during the fall, find the bottom speed.

Question 4 - Solution

A fixed hinge does no work while a rod falls

(a) Derivation and calculation

The center drops by $L/2$, and the fixed frictionless hinge does no work: $$\Delta U=-Mg\frac L2,\qquad \frac12I\omega^2=Mg\frac L2,\qquad I=\frac{ML^2}{3}.$$ $$\boxed{|\omega|=\sqrt{\frac{3g}{L}}=\sqrt{29.4}=5.422\ \mathrm{rad/s}}.$$ The signed angular velocity is clockwise for a rod initially extending to the right.

(b) Result and interpretation

If the brake removes half the released gravitational energy, $$K_f=\frac12\left(Mg\frac L2\right)=\frac{MgL}{4}.$$ $$\boxed{|\omega|=\sqrt{\frac{2K_f}{I}} =\sqrt{\frac{3g}{2L}}=\sqrt{14.7}=3.834\ \mathrm{rad/s}}.$$ The speed decreases by a factor $1/\sqrt2$, not by a factor of one half.

Physical check

Mass cancels from both speeds because gravitational energy and inertia are proportional to mass.